Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12






Last updated at Jan. 7, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Transcript
Misc 15 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4π/3 . Given, Radius of sphere = r Let R be the radius of the cone and H be its height. Let β BOC = ΞΈ Now, AC = AO + OC H = r + r cos ΞΈ And, R = r sin ΞΈ H = r (1 + cos ΞΈ) We need to maximize volume of cone. Volume of the cone is V = 1/3 ππ ^2 π» V = 1/3 ππ^2 sin^2β‘"ΞΈ" r (1 + cos ΞΈ) V = 1/3 ππ^3 sin^2β‘"ΞΈ" (1 + cos ΞΈ) Finding ππ£/ππ ππ£/ππ = 1/3 ππ^3 [2 sinβ‘γ"ΞΈ" cosβ‘γ"ΞΈ" (1+cosβ‘γ"ΞΈ" )+sin^2β‘γ"ΞΈ" (βsinβ‘γ"ΞΈ" )γ γ γ γ γ ] ππ£/ππ = 1/3 ππ^3 (2 sinβ‘γ"ΞΈ" cosβ‘γ"ΞΈ" (1+cosβ‘γ"ΞΈ" )βsin^3β‘"ΞΈ" γ γ γ ) ππ£/ππ = 1/3 ππ^3 (2 sinβ‘γ"ΞΈ" cosβ‘γ"ΞΈ" (1+cosβ‘γ"ΞΈ" )βsin^3β‘"ΞΈ" γ γ γ ) ππ£/ππ = 1/3 ππ^3 sin ΞΈ (2 cosβ‘γ"ΞΈ" +γ2 cos^2γβ‘γ"ΞΈ" βsin^2β‘"ΞΈ" γ γ ) ππ£/ππ = 1/3 ππ^3 sin ΞΈ (2 cosβ‘γ"ΞΈ" +γ2 cos^2γβ‘γ"ΞΈ" β(1βcos^2β‘"ΞΈ" γ γ)) ππ£/ππ = 1/3 ππ^3 sin ΞΈ (3 cos^2β‘γ"ΞΈ" +γ2 πππ γβ‘γ"ΞΈ" β1γ γ ) ππ£/ππ = 1/3 ππ^3 sin ΞΈ (3 cos^2β‘γ"ΞΈ" +γ3 πππ γβ‘π γ βcosβ‘π β1) ππ£/ππ = 1/3 ππ^3 sin ΞΈ (3 γπππ γβ‘"ΞΈ" (cosβ‘π+1)β1(cosβ‘π+1)) ππ£/ππ = 1/3 ππ^3 sin ΞΈ (3 πππ β‘γ"ΞΈ" β1γ )(πππ β‘γ"ΞΈ" +1γ ) Putting ππ£/ππ = 0 1/3 ππ^3 sin ΞΈ (πππ β‘γ"ΞΈ" +1γ ) (3 πππ β‘γ"ΞΈ" β1γ ) = 0 sin ΞΈ (cos ΞΈ + 1) (3 cos "ΞΈ" β 1) = 0 π¬π’π§ ΞΈ = 0 ΞΈ = 0Β° ΞΈ cannot be 0Β° for cone ππ¨π¬ ΞΈ + 1 = 0 cos ΞΈ = β1 For cone, 0Β° < ΞΈ < 90Β° & cos ΞΈ is negative in II & III quadrant. So cos ΞΈ = β1 is not possible π ππ¨π¬ ΞΈ β 1 = 0 cos ΞΈ = 1/3 ΞΈ = cosβ1 1/3 cos ΞΈ = 1/3 is possible So, πππ β‘"ΞΈ" = 1/3 Thus, H = r (1 + cos ΞΈ) H = r ("1 + " 1/3) H = 4π/3 This value is either maxima or minima. So, finding (π^2 π)/(ππ^2 ) ππ/ππ = (1/3 ππ^3 sinβ‘π (3 cos2 π + 2 cosβ‘πβ 1)) = 1/3 ππ^3 [cosβ‘π (3 cos^2β‘π+2 cosβ‘πβ1)+sinβ‘π (6 cosβ‘π (βπ ππ)β2 sinβ‘π )] = 1/3 ππ^3 [3 cos^3β‘π+2 cos^2β‘πβcosβ‘π+sinβ‘π (β6 cosβ‘π sinβ‘πβ2 sinβ‘π )] = 1/3 ππ^3 [3 cos^2β‘π+2 cos^2β‘πβcosβ‘πβ6 sin^2β‘π cosβ‘πβ2 sin^2β‘π ] Now, cosβ‘π=1/3 And sin2 "ΞΈ" = 1 β cos2 "ΞΈ" = 1 β (1/3)^2= 1 β 1/9 = 8/9 Putting values in (π^2 π)/(ππ^2 ) (π^2 π)/(ππ^2 ) = 1/3 ππ^3 [3(1/3)^3+2(1/3)^2β1/3β6(1/3)(8/9)β2(8/9)] = 1/3 ππ^3 [3(1/27)+2(1/9)β1/3β6(8/27)β2(8/9)] = 1/3 ππ^3 [1/9+ 2/9β1/3β16/9β16/9] = 1/3 ππ^3 [(β32)/9] = (β32ππ^3)/27 Thus, (π^2 π£)/(ππ^2 ) < 0 H = ππ/π Hence proved
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